回答編集履歴
2
ビット処理を使うコードを追加
test
CHANGED
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@@ -195,3 +195,81 @@
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}
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```
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**追記2**
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こんなやり方もあります。
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```Java
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class Test {
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public static void main(String[] args) {
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int num1[] = new int[10];
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int num2[] = new int[10];
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int bit1 = setRandomAndBit(num1);
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int bit2 = setRandomAndBit(num2);
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System.out.print("配列1:"); print(num1);
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System.out.print("配列2:"); print(num2);
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System.out.println("共通の数: "); print2(bit1 & bit2);
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System.out.println("どちらかの数: "); print2(bit1 ^ bit2);
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}
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static int setRandomAndBit(int[] a) {
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int bit = 0;
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for (int i = 0; i < 10; i++) {
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a[i] = (int) (Math.random() * 10) + 1;
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bit |= 1 << a[i];
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}
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return bit;
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}
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static void print(int[] a) {
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for (int i = 0; i < 10; i++)
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System.out.print(" " + a[i]);
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System.out.println();
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}
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static void print2(int bit) {
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for (int i = 1; i <= 10; i++)
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if ((bit >> i & 1) != 0) System.out.print(" " + i);
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System.out.println();
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}
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}
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```
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1
HashSet を使わないコードを追加
test
CHANGED
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@@ -87,3 +87,111 @@
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87
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どちらかの数: [2, 4, 6, 8, 10]
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```
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**追記**
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```Java
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class Test {
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public static void main(String[] args) {
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int num1[] = new int[10];
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int num2[] = new int[10];
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int same[] = new int[10];
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int side[] = new int[10];
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setRandom(num1);
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setRandom(num2);
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int n = 0, m = 0;
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for (int i = 0; i < 10; i++)
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if (find(num2, num1[i]))
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n = add(same, n, num1[i]);
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else
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m = add(side, m, num1[i]);
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for (int i = 0; i < 10; i++)
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if (!find(num1, num2[i]))
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m = add(side, m, num2[i]);
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System.out.print("配列1:"); print(num1, 10);
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System.out.print("配列2:"); print(num2, 10);
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System.out.print("共通の数:"); print(same, n);
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System.out.print("どちらかの数:"); print(side, m);
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}
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static void print(int[] a, int n) {
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for (int i = 0; i < n; i++)
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System.out.print(" " + a[i]);
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System.out.println();
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}
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static void setRandom(int[] a) {
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for (int i = 0; i < 10; i++)
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a[i] = (int) (Math.random() * 10) + 1;
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}
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static boolean find(int[] a, int e) {
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for (int i = 0; i < 10; i++)
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if (a[i] == e) return true;
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return false;
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}
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static int add(int[] a, int n, int e) {
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for (int i = 0; i < n; i++)
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if (a[i] == e) return n;
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a[n] = e;
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return n + 1;
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}
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}
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196
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+
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197
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+
```
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